Problem: Solve for $x$ : $6x^2 + 12x - 18 = 0$
Solution: Dividing both sides by $6$ gives: $ x^2 + {2}x {-3} = 0 $ The coefficient on the $x$ term is $2$ and the constant term is $-3$ , so we need to find two numbers that add up to $2$ and multiply to $-3$ The two numbers $3$ and $-1$ satisfy both conditions: $ {3} + {-1} = {2} $ $ {3} \times {-1} = {-3} $ $(x + {3}) (x {-1}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x + 3) (x -1) = 0$ $x + 3 = 0$ or $x - 1 = 0$ Thus, $x = -3$ and $x = 1$ are the solutions.